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If I have a typical Drude dielectric permittivity:

$$\epsilon(\omega) = \epsilon_{\infty} - \frac{\omega_p^2 \tau}{\omega^2\tau + i\omega}$$

Now, decomposing $\Re{(\epsilon(\omega))}$ and $\Im{(\epsilon(\omega))}$:

$$\Re{(\epsilon(\omega))}=-\frac{\omega_p^2}{\omega^2+1/\tau^2} + \epsilon_\infty$$ $$\Im{(\epsilon(\omega))} = \frac{\omega^2_p}{\omega^3 \tau+1/\tau}$$

Now, the imaginary part is the important one, because it relates the permittivity to conductivity (which is what I want).

$$\epsilon_{\text{im}}(\omega)= \frac{\sigma(\omega)}{\omega \epsilon_0}$$

So, why does equating the imaginary part and the above equation not yield the correct answer?

$$\sigma(\omega)=\frac{\omega_p^2\epsilon_0}{\omega^2\tau+1/\tau}$$

Should be:

$$\sigma_D(\omega) = \frac{\epsilon_0\omega_p^2 \tau}{1-i\omega\tau}$$

Where am I going wrong?

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2 Answers 2

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You’re going wrong by considering the optical conductivity to be real only. Conductivity is complex just as permittivity is! It can be related to epsilon through Ampere’s Law: $$\nabla\times H = J + \frac{dD}{dt}$$ For time-harmonic fields, we get $$\nabla\times H = (\sigma - i \omega \epsilon)E.$$ So the complex relative permittivity is related to the complex conductivity as $$\epsilon_r=\frac{\sigma}{i \omega \epsilon_0} - 1.$$ This is general (although there might be a sign discrepancy depending on your phasor convention) and, depending on your mood, can be taken as a matter of definition. If you leave out the imaginary part of $\sigma$, then you lose half of the information contained in the Drude model!

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  • $\begingroup$ So, given the complex relative permittivity and complex conductivity, how can I relate the two such that my last equation appears? $\endgroup$
    – smollma
    Nov 3, 2020 at 12:40
  • $\begingroup$ @smollma Try replacing the 1 in my expression with $\epsilon_\infty$ and possibly taking the complex conjugate. Let me know how it goes! $\endgroup$
    – Gilbert
    Nov 3, 2020 at 13:03
  • $\begingroup$ are you sure the last expressio nyou have is correct? Please see physics.stackexchange.com/q/591278 your 1 has a sign that would indicate a mismatched time harmonic dependence? $\endgroup$
    – smollma
    Nov 5, 2020 at 17:14
  • $\begingroup$ @smollma I’m not sure; feel free to proofread! My larger point is about recognizing the conductivity as complex. Does that make sense? $\endgroup$
    – Gilbert
    Nov 5, 2020 at 17:17
  • $\begingroup$ Yes, I understand that the conductivity should be complex. I was trying to equate the imaginary part of the conductivtiy and imag. part of permittivity (ignoring $i$) to use this to swap between the two functions. I can't seem to do it though. So: $$\epsilon_{\infty} +i \frac{\sigma(\omega)}{\omega \epsilon_0} = \epsilon_{\infty} - \frac{\omega_p^2 \tau}{\omega^2 \tau +i\omega}$$ This leaves me with: $$\sigma(\omega) = -i 2 \epsilon_{\infty}\omega\epsilon_0 - \frac{\omega_p^2\omega\epsilon_0\tau}{j(\omega^2\tau +i\omega)}$$ Assume $\epsilon_\infty$ is 1? $\endgroup$
    – smollma
    Nov 5, 2020 at 17:25
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Not really answering your question. However, if the only thing you are really wanting, is a derivation of the complex conductivity. You should start off with the drude model of conductivity. Namely, the differential equation

$m\frac{d^2\vec{x}}{dt^2} = q\vec{E_{0}}e^{-i\omega t} - \frac{m}{T} \frac{d\vec{x}}{dt}$

This models a single electron, moving at an instantaneous velocity $ \frac{d\vec{x}}{dt}$

Substituting this definition of instantaneous velocity of a single electron into the definition of current density. Acts as though this velocity is a VELOCITY FIELD, aka , at a SINGLE point in space, the velocity of an electron follows the instantaneous velocity of an electron moving under the differential equation above

The complex conductivity is a "Steady state solution" to this differential equation, so technically isn't accurate for low t values.

The way you solve this is via substitution in the form of a complex exponential. To find the transient solution, use the same substitution on the homogenous equation, and add it with the steady state solution

Drude models the E field as a complex number to make solving via substitution easier. THE ACTUAL CONDUCTIVITY is the REAL part

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